(3x+2)(4-x)=3

Solution for (3x+2)(4-x)=3 equation:

(3x+2)(4-x)=3

We move all terms to the left:

(3x+2)(4-x)-(3)=0

We add all the numbers together, and all the variables

(3x+2)(-1x+4)-3=0

We multiply parentheses ..

(-3x^2+12x-2x+8)-3=0

We get rid of parentheses

-3x^2+12x-2x+8-3=0

We add all the numbers together, and all the variables

-3x^2+10x+5=0

a = -3; b = 10; c = +5;
Δ = b2-4ac
Δ = 102-4·(-3)·5
Δ = 160
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-4\sqrt{10}}{2*-3}=\frac{-10-4\sqrt{10}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+4\sqrt{10}}{2*-3}=\frac{-10+4\sqrt{10}}{-6}
$